Why does the preservation of transition probabilities imply the preservation of all quantum probabilities?

I have a question about symmetries in quantum mechanics. Let $H$ be a Hilbert space, and $\mathbb{P}H$ the corresponding projective Hilbert (ray) space. In quantum mechanics, a symmetry is usually defined as a map $\mathcal{S}:\mathbb{P}H\rightarrow\mathbb{P}H$ which preserves the transition probabilities $$T(\bf{\Psi},\bf{\Phi}):=|\langle\psi,\phi\rangle|^{2}.$$ Here $\psi$ and $\phi$ denote arbitrary vectors contained in the rays $\bf{\Psi}$ and $\bf{\Phi}$ respectively. Thus, a symmetry should satisfy $$T(\mathcal{S}\bf{\Psi},\mathcal{S}\bf{\Phi})=T(\bf{\Psi},\bf{\Phi}).$$ Wigner's theorem then states that $\mathcal{S}$ lifts to $H$ as a unitary or antiunitary operator $U$.

Now let $P_{O}$ denote the projection-valued measure associated to a self-adjoint operator $O$ by the spectral theorem. If the system is described by the ray $\bf{\Psi}$, then the probability that a measurement of $O$ will produce a value in the Borel-measurable subset $V\subset\mathbb{R}$ is given by the Born probability $$\text{Prob}_{{\bf{\Psi}},O}(V):=\langle\psi,P_{O}(V)\psi\rangle$$ I would be very grateful if anyone could explain to me why the preservation of transition probabilities would necessarily imply the preservation of all such Born probabilities. I haven't seen a discussion of this point anywhere, which makes me think I might be missing something obvious.

• I'm not sure what your question is. $\mathrm{Prob}_{\Psi,O}(V) = T(\psi,P_O(V)\psi)$, i.e. the Born probabilities are transition probabilities, hence preserved. – ACuriousMind Jan 7 '16 at 21:24
• The action of the symmetry can be transposed on observables, and it turns out to be $O\mapsto\mathcal SO = UOU^*$. When you compute a transformed expectation value you are changing both the state and the observable according to $\psi\mapsto U\psi$ and $O\mapsto UOU^*$, whence the invariance. – Phoenix87 Jan 7 '16 at 21:27
• @Phoenix87 OK I see, so we need Wigner's theorem. Thank you both. So is it true that $P_{UOU^{*}}(V)=UP_{O}(V)U^{*}$? If so, is there a simple proof? My functional analysis isn't up to much I'm afraid :). – zander89 Jan 7 '16 at 21:49
• for the function $x^2$ you have $(UOU^*)^2 = UO^2U^*$, hence $p(UOU^*)=Up(O)U^*$ for any polynomial $p$. By Stone-Weierstrass the same is then true for continuous functions. To get the characteristic function of a Borel set simply take pointwise limits of continuous functions. – Phoenix87 Jan 7 '16 at 21:55
• @Phoenix87 Great, cheers. – zander89 Jan 7 '16 at 22:23